A particle moves along a straight line such t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The displacement is given by $s = t^3 - 3t^2 + 2$.The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The displacement is given by $s = t^3 - 3t^2 + 2$.The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 6t$.The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = 6t - 6$.Set the acceleration to zero to find the time $t$:$6t - 6 = 0 \implies t = 1 \, s$.Now,substitute $t = 1 \, s$ into the displacement equation to find the displacement at that instant:$s = (1)^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0 \, m$.

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 3t^2 + 2) \, m$. The displacement when the acceleration becomes zero is ........ $m$.

Similar Questions

For the acceleration-time $(a-t)$ graph shown in the figure, the change in velocity of the particle from $t=0$ to $t=6 \, s$ is ........ $m/s$.

If $v = x^2 - 5x + 4$,find the acceleration of the particle when the velocity of the particle is zero.

The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k} \text{ m/s}$. The magnitude of acceleration at point $(1, 2, 4)$ is . . . . . . $\text{m/s}^2$.

Which physical quantity can be found by first differentiation and second differentiation of position vector?

$Assertion$ : Retardation is directly opposite to the velocity.
$Reason$ : Retardation is equal to the time rate of decrease of speed.

Vedclass Test Series

Exam Paper Generator

Online Exam Module